Instiki
Sandbox

Skip the Navigation Links | Home Page | All Pages | Recently Revised | Authors | Feeds | Export |

Table of Contents (Skip the Table of contents)

Instiki

Edit this sidebar

Markdown+itex2MML Sandbox

Play around below. Your changes will, periodically, be rolled back.

Some examples

(1)minw hp h+w rp r+w lp l\mathop{min} w_h p_h + w_r p_r + w_l p_l

Here’s an equation

(2)∫ −∞ ∞e −ax 2/2dx=2πa{\int_{-\infty}^\infty e^{-a x^2/2} \mathrm{d}x} = \sqrt{\frac{2\pi}{a}}

which we can later refer1 back to as (2).

Aligned equations:

(3)a+b =b+a a+(b+c) =(a+b)+c\begin{aligned} a+b &= b+a \\ a+(b+c) &= (a+b)+c \end{aligned}

The Dirac equation (boxed):

(iD+m)ψ=0\boxed{(i\slash{D}+m)\psi = 0}

Here’s the table of Clifford2 algebras over ℝ:

j012345678
𝒞ℓ j −ℝℂℍℍ⊕ℍℍ(2)ℂ(4)ℝ(8)ℝ(8)⊕ℝ(8)ℝ(16)
𝒞ℓ j +ℝℝ⊕ℝℝ(2)ℂ(2)ℍ(2)ℍ(2)⊕ℍ(2)ℍ(4)ℂ(8)ℝ(16)

where the generators of 𝒞ℓ j ± satisfy

γ iγ j+γ jγ i=±2δ ij\gamma_i\gamma_j +\gamma_j \gamma_i =\pm 2\delta_{i j}

and 𝒞ℓ n+8 ±=𝒞ℓ n ±⊗ℝ(16).

(4)lim n→∞−1∑ k=1 n1k 2=π 26\lim_{n \to \infty-1} \sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6}

More Examples

(5)∇×E⇀=−∂B⇀∂t\nabla \times \vec{E} = - \frac {\partial \vec{B}}{\partial t}
(6)∮B⋅dl=μ 0I enc\oint \mathbf{B}\cdot \mathrm{d}\mathbf{l} = \mu_0 I_\text{enc}

H 1(𝒵,𝒪(−k)) Let G=(V,E) be a graph, with w:V→[0,1] a weight function.

(7){Q i,Q j}=δ ijℋ.\{Q_i, Q_j\} = \delta_{ij}\mathcal{H}.

Theorems

Definition

Let H be a subgroup of a group G. A left coset of H in G is a subset of G that is of the form xH, where x∈G and xH={xh:h∈H}.

Similarly a right coset of H in G is a subset of G that is of the form Hx, where Hx={hx:h∈H}.

Lemma

Let H be a subgroup of a group G, and let x and y be elements of G. Suppose that xH∩yH is non-empty. Then xH=yH.

Proof

Let z be some element of xH∩yH. Then z=xa for some a∈H, and z=yb for some b∈H. If h is any element of H then ah∈H and a −1h∈H, since H is a subgroup of G. But zh=x(ah) and xh=z(a −1h) for all h∈H. Therefore zH⊂xH and xH⊂zH, and thus xH=zH. Similarly yH=zH, and thus xH=yH, as required.

Lemma

Let H be a finite subgroup of a group G. Then each left coset of H in G has the same number of elements as H.

Proof

Let H={h 1,h 2,…,h m}, where h 1,h 2,…,h m are distinct, and let x be an element of G. Then the left coset xH consists of the elements xh j for j=1,2,…,m. Suppose that j and k are integers between 1 and m for which xh j=xh k. Then h j=x −1(xh j)=x −1(xh k)=h k, and thus j=k, since h 1,h 2,…,h m are distinct. It follows that the elements xh 1,xh 2,…,xh m are distinct. We conclude that the subgroup H and the left coset xH both have m elements, as required.

Theorem

(Lagrange’s Theorem). Let G be a finite group, and let H be a subgroup of G. Then the order of H divides the order of G.

Proof

Each element x of G belongs to at least one left coset of H in G (namely the coset xH), and no element can belong to two distinct left cosets of H in G (see Lemma 1). Therefore every element of G belongs to exactly one left coset of H. Moreover each left coset of H contains ∣H∣ elements (Lemma 2). Therefore ∣G∣=n∣H∣, where n is the number of left cosets of H in G. The result follows.

Corollary

Let x be an element of a finite group G. Then the order of x divides the order of G.

Theorem

Let f:Δ⟶Δ, where Δ={z∈ℂ:∣z∣<1}, be analytic with a∈Δ. Then

∣f(z)−f(a)1−f(a)¯f(z)∣≤∣z−a1−a¯z∣\left\vert\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}\right\vert\le \left\vert\frac{z-a}{1-\overline{a}z}\right\vert

for all ∣z∣≤1 and

∣f′(a)∣1−∣f(a)∣ 2≤11−∣a∣ 2.\frac{\vert f'(a)\vert}{1-\vert f(a)\vert^2}\le \frac{1}{1-\vert a \vert^2}.

Furthermore, equality holds iff f realizes a conformal mapping of Δ onto itself.

Proof

Let w=z−a1−a¯z and put ϕ(w)=f(z)−f(a)1−f(a)¯f(z). Define for absb<1 C b(z)=z−b1−b¯z. All conformal maps from Δ to itself, sending b to 0, are of the form C b(z)e iγ for γ∈[0,2π]. In this notation, ϕ(w)=C f(a)∘f∘C a −1(w), where C a −1 is the inverse of C a as a function. Note that C a(z) is conformal, so it has an inverse. It is clear that ϕ(0)=C f(a)∘f∘C a −1(0)=C f(a)(f(a))=0. Since C a −1:Δ⟶Δ and f:Δ⟶Δ and C f(a):Δ⟶Δ, then ∣ϕ(w)∣<1 for ∣w∣<1. Applying Schwarz’s lemma, we obtain ∣ϕ(w)∣≤∣w∣ for ∣w∣≤1. Furthermore, if equality holds, then f(z)=e iγ′z for γ′∈[0,2π]. Therefore,

(8)∣f(z)−f(a)1−f(a)¯f(z)∣≤∣z−a1−a¯z∣\left\vert\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}\right\vert\le \left\vert\frac{z-a}{1-\overline{a}z}\right\vert

for all ∣z∣≤1. Rearranging, we obtain

∣f(z)−f(a)z−a∣≤∣1−f(a)¯f(z)1−a¯z∣.\left\vert\frac{f(z)-f(a)}{z-a}\right\vert\le\left\vert\frac{1-\overline{f(a)}f(z)}{1-\overline{a}z}\right\vert.

If we take the limit as z tends to a, we obtain

∣f′(a)∣≤∣1−∣f(a)∣ 21−∣a∣ 2∣=1−∣f(a)∣ 21−∣a∣ 2,\left\vert f'(a)\right\vert \le \left\vert\frac{1-\vert f(a)\vert ^2}{1-\vert a \vert^2}\right\vert=\frac{1-\vert f(a)\vert^2}{1-\vert a \vert^2},

or

∣f′(a)∣1−∣f(a)∣ 2≤11−∣a∣ 2.\frac{\vert f'(a)\vert}{1-\vert f(a)\vert^2}\le \frac{1}{1-\vert a \vert^2}.

As said above, if equality holds in (8), then Schwarz’s lemma tells us that ϕ(w)=e iγ′w. Thus, ϕ(w)=C f(a)∘f∘C a −1(w)=e iγ′w, so f(z)=C f(a) −1(e iγ′C a(z)). Since e iγ′C a(z) is conformal, C f(a) −1, the inverse function of C f(a), is conformal, and a composition of conformal maps is conformal, then f is a conformal map of Δ onto itself. Conversely, if f is a conformal map of Δ onto itself, then ϕ(w)=C f(a)∘f∘C a −1(w)=e iγC b(w), since a composition of conformal maps is conformal and because all conformal maps from Δ onto itself are of the form e iγC b(w). We also know that ϕ(0)=0, so b=0. Therefore,

ϕ(w)=e iγC 0(w)=e iγw⇔∣ϕ(w)∣=∣w∣⇔∣f(z)−f(a)1−f(a)¯f(z)∣=∣z−a1−a¯z∣\phi(w)=e^{i\gamma}C_0(w)=e^{i\gamma}w \Leftrightarrow \vert\phi(w)\vert=\vert w \vert \Leftrightarrow \left\vert\frac{f(z)-f(a)}{1-\overline{f(a)}f(z)}\right\vert=\left\vert\frac{z-a}{1-\overline{a}z}\right\vert

for all ∣z∣≤1. In sum, equality holds in (8) iff f is a conformal map from Δ to itself.

Remark

Someone needs to code \abs, i.e. \left\vert # \right\vert. This is terribly annoying. Also, when using $$, why must one enter a line before and after.

SVG graphics, created in SVG-Edit (Instiki 0.19 feature):

Layer 1 E Σ E_\Sigma E Σ out E_{\Sigma_{\text{out}}} E in E_{\text{in}} [] Σ , X ] [\Sigma, X] [] Σ out , X ] [\Sigma_{\text{out}}, X] [] Σ in , X ] [\Sigma_{\text{in}}, X] Bord ( X ) Bord(X) V V E out E_{\text{out}} in \text{in} out \text{out} exp ( S ∇ ) ∣ Σ in \exp(S_\nabla)\vert_{\Sigma_{\text{in}}} exp ( S ∇ ) \exp(S_\nabla) exp ( S ∇ ) ∣ Σ out \exp(S_\nabla)\vert_{\Sigma_{\text{out}}} E in E_{\text{in}}
Commutative cube Layer 1 A 0 A_0 C 0 C_0 A 1 A_1 C 1 C_1 B 0 B_0 D 0 D_0 B 1 B_1 D 1 D_1
Box diagram W + W^+ W − W^- s ¯ \overline{s} d ¯ \overline{d} Layer 1 s s d d u , c , t u,\, c,\, t u , c , t u,\, c,\, t


K 0K¯ 0 Mixing

Yet More examples

  • SVG:
  • Animated SVG
Example anim01 - demonstrate animation elements ∫ −∞ ∞e −ax 2dx=πa
  • Complicated commutative diagrams (equations in SVG)
Complicated commutative diagram, realized in SVG 1 1 1 1 Id Id A B ρ H H K K′ ϕ 1 ϕ 2 N A N B N A ∨ N B ∨
  • SVG in equations.

In SU(3), Rank-2 Symmetric Tensor Representation ⊗ Fundamental Representation = Adjoint Representation ⊕ Rank-3 Symmetric Tensor Representation .

  • Cases:
r a+1={0 with prob.exp(−θr a) max{δr a,z} with prob.1−exp(−θr a) r_{a+1} = \begin{cases} 0 & \text{with prob.}\quad \exp(-\theta r_a) \\ \max \lbrace \delta r_a, z \rbrace & \text{with prob.}\quad 1 - \exp(-\theta r_a) \\ \end{cases}
  • Stretchy Brackets:
q a(z)=σ a −1exp[−γ+zσ a]q_a(z) = \sigma_a^{-1} \exp{\left[ -\frac{\gamma + z}{\sigma_a} \right]}
  • Linearity of Quadrature Rules
∑ i=1 N(αf(x i)+βg(x i))w i=α∑ i=1 Nf(x i)w i+β∑ i=1 Ng(x i)w i\sum_{i = 1}^N {\left( {\alpha f(x_i ) + \beta g(x_i )} \right)w_i } = \alpha \sum_{i = 1}^N {f(x_i )w_i } + \beta \sum_{i = 1}^N {g(x_i )w_i }
  • Linearity of Integrals
∫ a b(αf(x)+βg(x))dx=α∫ a bf(x)dx+β∫ a bg(x)dx{\int_a^b {\left( {\alpha f(x)\, + \beta g(x)} \right)dx = } \alpha \int_a^b {f(x)\,dx} + \beta \int_a^b {g(x)\,dx} }

  • Can we talk about x i 2 inline? What about ∫ a bx 2dx? Inline fractions x−x 2x 1−x 2?

  • Big fractions

p 3(x)=(12)(x−12)(x−34)(x−1)(14−12)(14−34)(14−1)+(12)(x−12)(x−34)(x−1)(14−12)(14−34)(14−1)+(12)(x−12)(x−34)(x−1)(14−12)(14−34)(14−1)+(12)(x−12)(x−34)(x−1)(14−12)(14−34)(14−1)p_3 (x) = \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}}
  • Diagram
P 1(Y) → P 1(X) ↓ ⇓ ∼ ↓ T′ → T\begin{matrix} P_1(Y) &\to& P_1(X) \\ \downarrow &\Downarrow^\mathrlap{\sim}& \downarrow \\ T' &\to& T \end{matrix}
(9) A_n Quiver v1 v2 vn1 ≡(U(k) n 1,{v i})\array{\arrayopts{\align{center}} \begin{svg} <svg xmlns="http://www.w3.org/2000/svg" width="10em" height="10em" viewBox="0 0 120 120"> <desc>A_n Quiver</desc> <g fill="none" stroke="black" stroke-width="1"> <path d="M 60 25 l 25 20 l 0 30 l -25 20 l -25 -20 l 0 -30 z" /> <g stroke-dasharray="2"> <path d="M 60 0 l 0 25" /> <path d="M 85 45 l 25 -20" /> <path d="M 85 75 l 25 20" /> <path d="M 60 95 l 0 25" /> <path d="M 35 75 l -25 20" /> <path d="M 35 45 l -25 -20" /> </g> </g> <g fill="red" stroke="none"> <circle cx="60" cy="25" r="4" /> <circle cx="85" cy="45" r="4" /> <circle cx="85" cy="75" r="4" /> <circle cx="60" cy="95" r="4" /> <circle cx="35" cy="75" r="4" /> <circle cx="35" cy="45" r="4" /> </g> <foreignObject x="45" y="0" width="16" height="20"><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><msub><mi>v</mi><mn>1</mn></msub></math></foreignObject> <foreignObject x="83" y="15" width="16" height="20"><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><msub><mi>v</mi><mn>2</mn></msub></math></foreignObject> <foreignObject x="5" y="25" width="20" height="25"><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><msub><mi>v</mi><mrow><msub><mi>n</mi><mn>1</mn></msub></mrow></msub></math></foreignObject> </svg> \end{svg} } \equiv \left({U(k)}^{n_1},\{v_i\}\right)

“This is my text”, says Anymouse.

Ruby code example:



    class Person
      attr_reader :name, :age
      def initialize(name, age)
        @name, @age = name, age
      end
      def <=>(person) # Comparison operator for sorting
        @age <=> person.age
      end
      def to_s
        "#@name (#@age)"
      end
    end
 
    group = [
      Person.new("Bob", 33), 
      Person.new("Chris", 16), 
      Person.new("Ash", 23) 
    ]
 
    puts group.sort.reverse

123

  1. You can also refer to it as (2). Chacun à son goût!. ↩

  2. For more information, see Wikipedia. ↩

Revised on May 22, 2013 04:16:04 by Anonymous Coward (82.209.67.209)
Edit | Back in time (344 revisions) | See changes | History | Views: Print | TeX | Source | Linked from: HomePage, contents, Installation, TODO, Jacques Distler, Sandbox, Upgrading, News, Installing on Shared Hosts, AnonymousCoward, S5, SVG, File Uploads, Housekeeping, Proxying, Bazaar, Categories, Syntax, Acknowledgements, Security, Migrating to MySQL, Edit Web, Launch at Startup, Browsers, Mongrel, Theorems, AccessKeys, Search, Known Bugs, BlahTeX, Commandline Options, Logo, Markup Choices, Run on Port 80, itex Endpoint, Hosted Services